Question

The solution of ${\sin }^{-1}x-{\sin }^{-1}2x=\pm \frac{\pi }{3}$ is

• A. $\pm \frac{1}{3}$
• B. $\pm \frac{1}{4}$
• C. $\pm \frac{\sqrt{3}}{2}$
• D. $\pm \frac{1}{2}$

Answer 1

Answer: (D)

$\pm \frac{1}{2}$

${\sin }^{-1}x-{\sin }^{-1}2x=\pm \frac{\pi }{3}$
$\Rightarrow {\sin }^{-1}x-{\sin }^{-1}(\pm \frac{\sqrt{3}}{2})={\sin }^{-1}2x$
$\Rightarrow {\sin }^{-1}[x\sqrt{1-\frac{3}{4}}-(\pm \frac{\sqrt{3}}{2}\sqrt{1-x^2})]={\sin }^{-1}2x$
$\Rightarrow \frac{x}{2}-(\pm \frac{\sqrt{3}}{2}\sqrt{1-x^2})=2x$
$\Rightarrow -(\pm \sqrt{3}\sqrt{1-x^2})=3x$
On squaring, both sides we get,
$3(1-x^2)=9x^2$
$\Rightarrow 1-x^2=3x^2$
$\Rightarrow 4x^2=1$
$\Rightarrow x=\pm \frac{1}{2}$